btw: as much as I love Chung, it's not an easy book to use if you're self-teaching

I often find the solutions hard to follow.

In this case, I misunderstood Chung's solution. The solution I **thought** he was using was in fact wrong...but getting balled up like that meant that I didn't manage to pick up on my 'careless error'...

(I'm using 'careless error' to mean that I was approaching the problem correctly but missed a step & then missed the fact that I'd missed a step.)

What method did he use? Mine worked, but took a long time. I found the area of the inner circle (9pi) first. The middle circle is 36pi - 9pi (because it has a 9pi-sized "hole"), or 27pi. The outer circle is 81pi - 36pi (area of the other two rings, which have already been accounted for), or 45pi. Then I added the area of the shaded parts together, 9pi + 45pi = 54pi, and divided by the total area (81pi) to determine what percentage of the darts fall in the shaded range.

Don't you mean: inner + outer = 6 (by which I assume you mean, the shaded area is 6 times the center circle); whole dartboard is 9 (i.e. 9 times the center circle); 6:9 = 16:24?

My approach, by the way, was to ignore factors of pi (since I knew we were interested in a ratio of areas, the pi's would all cancel out anyway). Outermost circle area is 81, middle circle area is 36, innermost circle area is 9; so shaded area is 81-36+9 = 54. As a fraction of total area, it's 54/81 = 2/3. (That's the steps where the pi's would have canceled if I had been keeping track of them.) So 2/3 of the darts would be expected to hit the shaded area.

If the inner circle has area A units, and the rings all have width equal to the radius of the inner circle, then the areas will be 1A, 3A, 5A, 7A, ... and a the probability of landing in a shaded region as a function of the number of rings (call it P(R)) will beP(1) = 1P(2) = 1/4P(3) = 2/3P(4) = 3/8P(5) = 3/5P(6) = 5/12P(7) = 4/7and there are all sorts of neat, accessible patterns to play with and explore.